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Let $ A $ be an $ \mathcal{H}^1$-measurable subset of $ \mathbb{R} $ and $ \gamma \colon A \subseteq \mathbb{R} \to \ell^\infty $ be a Lipschitz mapping with the Lipschitz constant $ L $. Also, assume that for all $ n \in \mathbb{N} $ and $ \mathcal{H}^1$-a.e. $ t \in A $, $$ \gamma_n'(t) = 0, $$ where $ \gamma_n $ is the $ n^{th} $ component of $ \gamma $.

I want to prove that $$ \lim_{h \to 0}{\frac{\|\gamma(t+h) - \gamma(t)\|_\infty}{\vert h \vert}} = 0, $$ at $ \mathcal{H}^1$-a.e. $ t \in A $.

Any suggestions or ideas are greatly welcomed. Thanks.

Edit: If $ A $ happens to be an interval, the proof is extremely easy because each component is going to be constant. The problem is $ A $ is not necessarily an interval.

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Is ${\cal H}^1$ one-dimensional Hausdorff measure? So this is just Lebesgue measure?

Then I think the answer is yes, specifically the desired limit is zero at every Lebesgue point of $A$. If $t$ is a Lebesgue point then for any $\epsilon > 0$ we can find $r > 0$ such that $\frac{\mu(A \cap I)}{\mu(I)} \geq 1-\epsilon$ for any interval $I$ centered at $t$ of length at most $2r$. Then any function which has Lipschitz number at most $L$ and whose derivative is zero on $A$ will satisfy $\frac{|f(t+h)-f(t)|}{|h|} \leq 2L\epsilon$ for all $|h| \leq r$. So all components will satisfy this inequality, hence you also get it when you take the sup norm.

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  • $\begingroup$ Thanks Nik. I just have a quick question. How did you conclude that $ \frac{\vert f(t+h) - f(t) \vert}{\vert h \vert} \leq L\epsilon $ for all $ \vert h \vert \leq r $? $\endgroup$
    – Axiom
    Commented Jan 18, 2014 at 23:07
  • $\begingroup$ And yes, $ \mathcal{H}^1 $ is just the 1-dimensional Hausdorff measure on $ \mathbb{R} $ which coincides with the Lebesuge measure on $ \mathbb{R} $. $\endgroup$
    – Axiom
    Commented Jan 18, 2014 at 23:10
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    $\begingroup$ Extend $f$ to all of ${\bf R}$ without increasing its Lipschitz number, then it is the integral of its derivative which has $L^\infty$ norm at most $L$, and you get the inequality by integrating its derivative. $\endgroup$
    – Nik Weaver
    Commented Jan 18, 2014 at 23:10
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    $\begingroup$ (If $f$ is complex-valued, then you may have to increase its Lipschitz number by at most a factor of $\sqrt{2}$ when extending it.) $\endgroup$
    – Nik Weaver
    Commented Jan 18, 2014 at 23:11
  • $\begingroup$ Ok, so I think there might be a small problem with your argument. The problem is when you extend $ f $ to the entire set of $ \mathbb{R} $, let's call this extension $ F $, then $ F(t+h) $ is not necessarily equal to $ f(t+h) $ because $ t+h $ is not in $ A $ necessarily. Hence, the estimate $ \frac{\vert f(t+h)-f(t)\vert}{\vert h \vert} \leq 2L\epsilon $ doesn't hold. Am I missing something here possibly? $\endgroup$
    – Axiom
    Commented Jan 20, 2014 at 23:22

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